Counting all the pairs, and why swapping the two sets gives something else

0:00Two sets. Nobody says what is in them. One has four members. The other has three. How many pairs can you build, taking the first entry from the first set and the second entry from the second? Twelve. And you can say that without knowing a single name. That is worth being suspicious about. A count you can state without looking is either a deep fact or a rule somebody told you to memorise.

0:29This one is neither. It is a description of how the pairs got made. So let us make them, and count what comes out. Take the first of the four. Offer it every one of the three. Three pairs. Now the second one. Three more. The third. Three more. The fourth. Three more. Four rows of three. Twelve. That is where the rule comes from. Not from multiplication arriving out of nowhere. From four helpings of three.

1:03Now check it properly. Take four names, and all sixteen sets you can make from them. That is two hundred and fifty-six ordered pairs of sets. Build every product by hand and count what is in it. All two hundred and fifty-six agree with the two counts multiplied. And in every one of them, each first entry gets the whole of the second set. Five hundred and twelve rows, all of them full.

1:34A count is only right if nothing got written down twice. So why can two of those twelve never be the same pair? Because a pair is settled by its two positions. Any two pairs from the grid come from a different row, or a different column, or both. A different row means different first entries. A different column, different second entries. Either way one position disagrees, so the pairs disagree.

2:04Take the four-by-four grid. Sixteen pairs. Compare every written pair against every other one. Two hundred and forty comparisons. Not one of them finds the same pair twice. There is one place where the writing and the counting come apart. Write a member of a factor down twice. Now the construction writes six pairs where there are four. It is the same product as before. The extra row was never a new row, and the count knows it.

2:38Now the warning everybody half-remembers. A set of three: a, b and c. A set of one: r. Cross them one way. a with r, b with r, c with r. Three pairs. Cross them the other way. r with a, r with b, r with c. Three pairs. How many appear on both lists? None. So the two products are different sets. Not slightly different. They share nothing at all.

3:07And they have exactly the same number of members. Three ones and one three are both three, and that is not a coincidence. It is the same rule read twice. The count survives the swap. The pairs do not. That needs separating carefully, because most people hear only half of it. Different sets is one claim. Different sizes is another. Only the first one is true. So measure it. Five names, thirty-two sets, one thousand and twenty-four ordered pairs of them.

3:44For every single one of the one thousand and twenty-four, the product and its reverse have the same count. And they are the same set only ninety-four times. Which ninety-four? Exactly the ones where the two sets are the same set, or one of them is empty. Nothing else, ever. Put that description beside the equality across all one thousand and twenty-four, and the two never once disagree. So swapping almost always changes the answer, and never changes the count.

4:20Almost always changes the answer. By how much? Here is something the warning does not tell you. Take what the two sets share. Cross that with itself. That is exactly the overlap between a product and its reverse. Exactly. Not roughly. On all one thousand and twenty-four, it lands on the nose. Watch it work. One, two, three against two, three, four. They share two and three. Two members, so the shared part crossed with itself holds four pairs.

4:54And the product and its reverse share four pairs, and those are the four. Two of the four have entries that differ, and those are the interesting ones. Each of those two sits in one product with its swap sitting in the other. If the two sets share nothing, the overlap is nothing. Two hundred and forty-three of the pairs share nothing, and all two hundred and forty-three have products that share nothing.

5:23Now break the rule on purpose. Everything so far assumed both sets can be counted. Keep the first set at three members and let the second one grow. Three. Six. Nine. Twelve. Every time you add one member, the product gains three more. Twelve steps in it is thirty-six, and it went up at every single step. Eleven steps, eleven increases. So if the second set cannot be counted, neither can the product.

5:55Here is the honest way to see that. Pick one first entry and hold it fixed. Now every member of the second set gives you a different pair. Twelve sizes checked, and every time, holding one first entry keeps every second entry apart. So the product is at least as big as the set you are crossing in. If that one is endless, so is the product. And then the rule says nothing at all, because it needed two numbers to multiply and it only has one.

6:30There is a case where that endlessness is the entire point. Take all the real numbers and cross them with themselves. Every pair of real numbers. A first one and a second one. That is not two number lines lying side by side. It is the plane. Every point of it. One point for each pair, and the pair is the point's two coordinates. A point of the plane needs exactly two numbers, which is exactly what a pair supplies.

7:02Do it once more and you get three positions instead of two. Three numbers. That is ordinary space. The counting rule has nothing to say about either of them. The construction still does. Three positions deserve a closer look. Take the set holding one and two. Build every triple. First from the set, second from the set, third from the set. Two choices, then two, then two. Eight. That is the same argument again, run three times instead of twice.

7:38Now a question the count cannot answer. Is a triple the same thing as a pair whose first entry is itself a pair? Bracket the first two together and you get eight objects. Bracket the last two together and you get eight objects. And there are eight triples. Three collections of eight. How many members do any two of them share? None. Not one, in any direction. The same count, three different sets, and only one of the three has three slots.

8:11Two more things a product seems to do. Three sets. One, two and three. Three and four. Four, five and six. What do the second and third share? Just the number four. Cross the first set with that. Three pairs. Now go the other way round. The first set crossed with the second holds six pairs. Crossed with the third, nine pairs. What do those two products share? Three pairs.

8:40Three and three, and they are the same three. So crossing seems to pass straight through what two sets share. Try it on everything the two hold between them. The second and third sets together hold three, four, five and six. Four members. The first set crossed with that is three times four. Twelve pairs. Other route. Take the two products, six pairs and nine pairs, and put them together.

9:10Six plus nine is fifteen. But three of them were counted twice, because those three sit in both. Fifteen take away three is twelve. Twelve and twelve. The same answer, and the arithmetic tells you why. So it passes through both-together as well. Now the part that matters. What has actually been shown? One choice of three sets. One. Take three names. There are eight sets you can make from them, so five hundred and twelve ordered triples.

9:45The check settled one of those and left five hundred and eleven untouched. Run the machine over all five hundred and twelve and both rules do hold, every time. But five hundred and twelve is still a number, and there are endlessly many triples of sets. And there is worse. Many of those checks are asleep. In two hundred and sixteen of the five hundred and twelve, the second and third sets share nothing at all.

10:15Both sides are empty, and the check confirms that empty is empty. So argue instead. Take any pair at all. It is on the left when its first entry is in the first set and its second entry is in both of the others. It is on the right when it is in both products, and that says the same two things. The same condition, so the same pairs. Every pair, and every choice of sets.

10:43Asked of every object of all five hundred and twelve triples, four thousand six hundred and eight questions, the two sides never once disagree. Now run the whole thing backwards. Somebody hands you four pairs. p with q, p with r, m with q, m with r. Which two sets were crossed? Collect the openings. p and m. Collect the closings. q and r. Two and two, and two times two is four, so nothing has gone missing.

11:18Cross them back and you get exactly what you were handed. But notice what you were told. You were told this was a product. Three names give nine pairs, and therefore five hundred and twelve collections you could make out of them. Only fifty of the five hundred and twelve are the product of anything. Four hundred and sixty-two are not. Here is one of those. Just two pairs. Its openings are two things and its closings are two things, so openings crossed with closings would give four pairs.

11:54It only has two. So it is not a product. Rebuilding is the test. If it does not rebuild, it never was one. One last one, and it is the count running backwards. A set crossed with itself has nine members. What is the set? Nine is three times three, and no other whole number squared is nine. So the set has three members. But that is all the count gives you.

12:23Six names on their own already allow twenty different three-member sets, and every one of the twenty has a self-product of nine. The count fixes the size and nothing else. So now take two of the nine pairs. Minus one with zero, and zero with one. Those two pairs mention three different numbers. Minus one, zero and one. Three numbers, and the set has room for exactly three. So that is the set, and there is no other.

12:55Nine pairs, two of them given, seven left to write out. And one last count worth keeping. A set of two crossed with a set of two holds four pairs, and there are sixteen different collections you could pick out of those four. That is what the next idea is going to be made of.